Past exam of the topics in applied mathematics course of the University of Oxford / 2011 / A / 2 / c

Set $Y=U^{q/2}$ and $k=2b/q$. The proposed profile with $a,b>0$ givesSubstituting into $U^{\prime \prime }+c{U}^{\prime }+U(1-U^q)=0$ and dividing in the interior by $U(1-Y)$ yieldsEquating the constant and $Y$ coefficients givesHence the exact generalized Fisher front isThe free positive $a$ translates the front. It connects one on the left to zero on the right and is strictly decreasing. At $q=1$, its speed is $5/\sqrt{6}$.

The explicit travelling wave speed increases with $q$:Its two tails change differently. Near the leading empty state, $U\sim a^{-2/q}{e}^{-kz}$ with $k=\sqrt{2/(q+2)}$ decreasing as $q$ grows: that tail lengthens. Near the filled state, $1-U\sim (2a/q)e^{bz}$, with $b=q/\sqrt{2(q+2)}$ increasing: that tail becomes steeper. Thus the change is not a uniform sharpening of the whole profile. For example the half-height slope is $-\frac{k}{2}(1-2^{-q/2})$.

For meaningful comparison, the figure aligns all fronts at $U(0)=1/2$ by choosing $a=2^{q/2}-1$. It displays the speed increase and the unequal changes of the two tails. Since $f^{\prime }(0)=1$ for every $q$, the minimum admissible speed remains two; the explicit family has $c(q)>2$ for $q>0$. The invariant-triangle proof above extends with boundary derivative $U^{q+1}\ge 0$.